There is a circular loop carrying current and another straight wire carrying current is placed along its axis. The straight wire

Options

A. will apply radial inward force on the circular loop.
B. will apply radial outward force on the circular loop.
C. will apply force radially inward or outward, depending on the direction of currents.
D. will not exert any force on the circular loop.

Explanation

To solve the problem of the interaction between a circular loop carrying current and a straight wire carrying current placed along its axis, we can follow these steps:

### Step 1: Understand the Configuration
We have a circular loop with a current \( I_1 \) flowing through it and a straight wire with current \( I_2 \) placed along the axis of the loop. We need to analyze the magnetic field produced by the straight wire and its effect on the circular loop.

**Hint:** Visualize the setup clearly, noting the direction of currents in both the loop and the wire.

### Step 2: Determine the Magnetic Field from the Straight Wire
A straight wire carrying current produces a magnetic field around it. The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I_2 \) is given by the formula:
\[ B = \frac{\mu_0 I_2}{2\pi r} \]
where \( \mu_0 \) is the permeability of free space.

**Hint:** Remember that the magnetic field lines form concentric circles around the wire.

### Step 3: Analyze the Magnetic Field at the Loop
Since the straight wire is along the axis of the circular loop, the magnetic field at any point on the loop will be directed tangentially to the loop. The direction of the magnetic field can be determined using the right-hand rule.

**Hint:** Use the right-hand rule to determine the direction of the magnetic field produced by the straight wire.

### Step 4: Calculate the Force on the Loop
The force \( dF \) on a small segment \( dL \) of the loop due to the magnetic field \( B \) is given by:
\[ dF = I_1 dL \times B \]
The angle between \( dL \) (tangential to the loop) and \( B \) (also tangential) is \( 180^\circ \) (or \( \pi \) radians), which means:
\[ \sin(180^\circ) = 0 \]
Thus, the force \( dF \) becomes zero:
\[ dF = I_1 dL \cdot B \cdot \sin(180^\circ) = 0 \]

**Hint:** Remember that the cross product is zero when the vectors are parallel or anti-parallel.

### Step 5: Conclude the Interaction
Since the force \( dF \) is zero for every segment of the loop, the total force \( F \) on the entire loop is also zero. Therefore, the straight wire does not exert any net force on the circular loop.

**Hint:** Consider the implications of having zero force at every point on the loop.

### Final Answer
The straight wire will not exert any force on the circular loop.

**Correct Option:** The straight wire will not exert any force on the circular loop.